RE: [EMHL] IMO training 2004 homework problem sets I, II, III

Dear Darij and Vladimir For the case that the quadrilateral is not convex from a sketch I concluded (perhaps this is not correct) that the diagonals BD, AC of

Message 1 of 12
, Apr 11, 2004

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Dear Darij and Vladimir

For the case that the quadrilateral is not convex from a sketch
I concluded (perhaps this is not correct) that the diagonals
BD, AC of the quadrilateral must be parallel and the point O
must lie on the line joining the mid points of the diagonals.

By a curious coincidence, I have found this
paper yesterday before receiving your mail! It
was indeed nice to see my conjecture verified
that the poristic locus (or "trajectory") of
the symmedian point is an ellipse. However,
concerning the proofs of the rather simple
poristic loci (i. e., those of the Nagel point,
nine-point center, orthocenter, centroid), I
prefer using the Feuerbach theorem rather than
calculating the distances using barycentric
coordinates.

Sincerely,
Darij Grinberg

Alexey.A.Zaslavsky

Dear Darij! ... You are right, for this points there are a synthetical proof. But without coordinates we could not prove the most interesting our result: the

Message 4 of 12
, Apr 12, 2004

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Dear Darij!

>
>By a curious coincidence, I have found this
>paper yesterday before receiving your mail! It
>was indeed nice to see my conjecture verified
>that the poristic locus (or "trajectory") of
>the symmedian point is an ellipse. However,
>concerning the proofs of the rather simple
>poristic loci (i. e., those of the Nagel point,
>nine-point center, orthocenter, centroid), I
>prefer using the Feuerbach theorem rather than
>calculating the distances using barycentric
>coordinates.
>

You are right, for this points there are a synthetical proof. But without
coordinates we could not prove the most interesting our result: the
trajectory of Gergonne point is the circle coaxial with the circumcircle and
the incircle.

Sincerely Alexey

ndergiades

Dear Darij and Vladimir sorry a correction. In my 9658 message ... ******** The point O is (0, -d) and not (-d , 0) Best regards Nikolaos Dergiades

Message 5 of 12
, Apr 12, 2004

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Dear Darij and Vladimir
sorry a correction. In my 9658 message
I wrote:

> For the case that the quadrilateral is not convex from a sketch
> I concluded (perhaps this is not correct) that the diagonals
> BD, AC of the quadrilateral must be parallel and the point O
> must lie on the line joining the mid points of the diagonals.
>
> For example if the points A,B,D,C are concyclic and the
> quadrilateral ABDC is isosceles trapezium AB=DC, AC || BD
> then the quadrilateral ABCD is not convex and if
> the cartesian coordinates are
> B(-a , 0) , D(a , 0) , O( -d, 0)
> where O is the point satisfying the "halfside" condition
> A(-p , q) , C( p , q), a, d, p, q > 0
> then p , q can be found from p/d = q/a
> (aa+dd)pp + 2ad(a-2d)p + ddaa = 0
>
> Special Case :
> d = 2a and p = 2a q = a or
> d = 2a and p = 2a/5 q = a/5

********

The point O is (0, -d) and not (-d , 0)

Best regards
Nikolaos Dergiades

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